Exercise 2.8 - Chapter 2 Integral Calculus I 12th Maths Guide Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated On May 15, 2024

By SaraNextGen

Ex $2.8$
I. Using the second fundamental theorem, evaluate the following:
Question 1.
$\int_0^1 e^{2 x} d x$
Solution:
$
\int_0^1 e^{2 x} d x=\left[\frac{e^{2 x}}{2}\right]_0^1=\frac{e^2}{2}-\frac{e^0}{2}=\frac{e^2-1}{2}
$
Question 2.
$
\int_0^{\frac{1}{4}} \sqrt{1-4 x} d x
$
Solution:
$
\begin{aligned}
\int_0^{\frac{1}{4}} \sqrt{1-4 x} d x & =\int_0^{\frac{1}{4}}(1-4 x)^{\frac{1}{2}} d x \\
& =\left[\frac{(1-4 x)^{\frac{3}{2}}}{\frac{3}{2}(-4)}\right]_0^{\frac{1}{4}}=\frac{\left[1-4\left(\frac{1}{4}\right)\right]^{\frac{3}{2}}}{-6}-\frac{[1-4(0)]^{\frac{3}{2}}}{-6}=0+\frac{1}{6}(1)^{\frac{3}{2}}=\frac{1}{6}
\end{aligned}
$
Question 3.
$
\int_1^2 \frac{x d x}{x^2+1}
$
Solution:
$
\begin{aligned}
\int_1^2 \frac{x d x}{x^2+1} & =\frac{1}{2} \int_1^2 \frac{2 x d x}{x^2+1} \\
& =\frac{1}{2} \int_1^2 \frac{d\left(x^2+1\right)}{x^2+1}=\frac{1}{2}\left[\log \left|x^2+1\right|\right]_1^2 \\
& =\frac{1}{2}\left[\log \left|2^2+1\right|-\log \left|1^2+1\right|\right] \\
& =\frac{1}{2}[\log 5-\log 2]
\end{aligned}
$

$
=\frac{1}{2} \log \left[\frac{5}{2}\right]\left\{\text { Using } \log a-\log b=\log \left(\frac{a}{b}\right)\right\}
$
Question 4.
$
\int_0^3 \frac{e^x d x}{1+e^x}
$
Solution:
$
\begin{aligned}
\int_0^3 \frac{e^x d x}{1+e^x} & =\int_0^3 \frac{d\left(1+e^x\right)}{1+e^x} \\
& =\left[\log \left|1+e^x\right|\right]_0^3=\log \left|1+e^3\right|-\log \left|1+e^0\right| \\
& =\log \left|1+e^3\right|-\log 2=\log \left[\frac{1+e^3}{2}\right]
\end{aligned}
$
Question 5.
$
\int_0^1 x e^{x^2} d x
$
Solution:
$
\int_0^1 x e^{x^2} d x=\frac{1}{2} \int_0^1 2 x e^{x^2} d x
$
Let $t=x^2$, Then $d t=2 x d x$
When $x=0, t=0 ; x=1, t=1$
So the integral becomes, $\quad \frac{1}{2} \int_0^1 e^t d t=\frac{1}{2}\left[e^t\right]_0^1=\frac{1}{2}[e-1]$

Question 6.
$
\int_1^e \frac{d x}{x(1+\log x)^3}
$
Solution:
Let $1+\log x=t, \quad$ Then $\frac{1}{x} d x=d t$
When $x=1, t=1 ; x=e, t=1+\log e=2$
$
=\int_1^2 \frac{d t}{t^3}=\int_1^2 t^{-3} d t=\left[\frac{t^{-2}}{-2}\right]_1^2
$
$
=\left[\frac{-1}{2 t^2}\right]_1^2=\frac{-1}{8}+\frac{1}{2}=\frac{3}{8}
$

Question 7.
$
\int_{-1}^1 \frac{2 x+3}{x^2+3 x+7} d x
$
Solution:
$
\begin{aligned}
\int \frac{23}{x+3 x+7} d x & =\int_{-1}^1 \frac{d\left(x^2+3 x+7\right)}{x^2+3 x+7} \\
& =\left[\log \left|x^2+3 x+7\right|\right]_{-1}^1 \\
& =\log |1+3+7|-\log |1-3+7| \\
& =\log 11-\log 5 \\
& =\log \left[\frac{11}{5}\right]
\end{aligned}
$
Question 8.
$
\int_0^{\frac{\pi}{2}} \sqrt{1+\cos x} d x
$
Solution:
$
\begin{aligned}
& \int_0^{\frac{\pi}{2}} \sqrt{1+\cos x} d x \\
& \text { We know } \cos 2 x=2 \cos ^2 x-1 \\
& \Rightarrow \quad \cos x=2 \cos ^2 \frac{x}{2}-1 \\
& \Rightarrow \quad 1+\cos x=2 \cos ^2 \frac{x}{2} \\
& \int_0^\pi \sqrt{2 \cos ^2 \frac{x}{2}} d x=\int_0^2 \sqrt{2} \cos \frac{x}{2} d x=\left[\frac{\sqrt{2} \sin \frac{x}{2}}{\frac{1}{2}}\right]_0^{\frac{\pi}{2}} \\
&=2 \sqrt{2} \sin \frac{\pi}{4}-2 \sqrt{2} \sin 0 \\
&=2 \sqrt{2}\left(\frac{1}{\sqrt{2}}\right)=2
\end{aligned}
$

Question 9.
$
\int_1^2 \frac{x-1}{x^2} d x
$
Solution:
$
\begin{aligned}
& \int_1^2 \frac{x-1}{x^2} d x=\int_1^2\left(\frac{x}{x^2}-\frac{1}{x^2}\right) d x \\
& =\int_1^2 \frac{1}{x} d x-\int_1^2 \frac{1}{x^2} d x \\
& =[\log |x|]_1^2-\left[\frac{-1}{x}\right]_1^2 \\
& =\log 2-\log 1+\frac{1}{2}-1=\log 2-\frac{1}{2} \\
& =\frac{1}{2}[2 \log 2-1]
\end{aligned}
$
II. Evaluate the following:
Question 1.
$
\int_1^4 f(x) d x \text { where } f(x)=\left\{\begin{array}{l}
4 x+3,1 \leq x \leq 2 \\
3 x+5,2 \end{array}\right.
$
Solution:
$
\begin{aligned}
\int_1^4 f(x) d x=\int_1^2 f(x) d x+\int_2^4 f(x) d x & =\int_1^4(4 x+3) d x+\int_2^4(3 x+5) d x \\
& =\left[\frac{4 x^2}{2}+3 x\right]_1^2+\left[\frac{3 x^2}{2}+5 x\right]_2^4 \\
& =2(4)+3(2)-2(1)-3(1)+\frac{3}{2}(16)+5(4)-\frac{3}{2}(4)-5(2) \\
& =8+6-2-3+24+20-6-10=37
\end{aligned}
$

Question 2.
$
\int_1^2 f(x) d x \text { where } f(x)=\left\{\begin{array}{l}
3-2 x-x^2, x \leq 1 \\
x^2+2 x-3,1 \end{array}\right.
$
Solution:
$
\begin{aligned}
\int_0^2 f(x) d x=\int_0^1 f(x) d x+\int_1^2 f(x) d x & =\int_0^1\left(3-2 x-x^2\right) d x+\int_1^2\left(x^2+2 x-3\right) d x \\
& =\left[3 x-x^2-\frac{x^3}{3}\right]_0^1+\left[\frac{x^3}{3}+x^2-3 x\right]_1^2 \\
& =3-1-\frac{1}{3}+0+0+0+\frac{8}{3}+4-6-\frac{1}{3}-1+3 \\
& =\frac{5}{3}+\frac{8}{3}-\frac{1}{3}=4
\end{aligned}
$
Question 3.
$
\int_{-1}^1 f(x) d x \text { where } f(x)=\left\{\begin{array}{l}
x, x \geq 0 \\
-x, x<0
\end{array}\right.
$
Solution:
$
\begin{aligned}
\int_{-1}^1 f(x) d x=\int_{-1}^0 f(x) d x+\int_0^1 f(x) d x & =\int_{-1}^0-x d x+\int_0^1 x d x \\
& =\left[\frac{-x^2}{2}\right]_{-1}^0+\left[\frac{x^2}{2}\right]_0^1=\frac{1}{2}+\frac{1}{2}=1
\end{aligned}
$
Question 4.
$
f(x)=\left\{\begin{array}{l}
c x, 0 0, \text { otherwise }
\end{array} \text { Find 'c'if } \int_0^1 f(x) d x=2\right.
$

Solution:
$
\text { Given, } \begin{aligned}
\int_0^1 f(x) d x & =2 \\
\text { ie., } \quad \int_0^1 c x d x & =2 \\
{\left[\frac{c x^2}{2}\right]_0^1 } & =2 \\
c\left(\frac{1}{2}\right) & =2 \quad \Rightarrow c=4
\end{aligned}
$